A) \[(1-sin2x).{{y}_{1}}\]
B) \[\left( 1+sin2x \right).{{y}_{1}}\]
C) \[\left( 1-cosx \right).y{{,}_{1}}\]
D) \[\left( 1-cos2x \right).{{y}_{1}}\]
Correct Answer: B
Solution :
[b] \[y={{e}^{tanx}}\] \[{{y}_{1}}={{e}^{\tan x}}.\sec x\] \[{{y}_{2}}={{e}^{\tan x}}.se{{c}^{4}}+{{e}^{\tan x}}.2.\sec x.tanx\] Now, \[={{e}^{\tan x}}\left( {{\sec }^{2}}x+\frac{2\tan x.se{{x}^{2}}x}{se{{x}^{2}}x} \right)\] \[={{e}^{\tan x}}.{{\sec }^{2}}x(1+2.\sin x.\cos x)\] \[={{e}^{\tan x}}.{{\sec }^{2}}x(1+\sin 2x)={{y}_{1}}(1+\sin 2x)\] Hence, option [b] is correct.
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